According to the graph of acceleration versus time, set the speed at times 4 and 15 s, if at time 1 s the speed is 3 \frac{m}{s}.
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<h3> Statement </h3>
<p>
$1.2.14.$ According to the graph of acceleration versus time, set the speed at times $4$ and $15$ s, if at time $1$ s the speed is $3$ $\frac{m}{s}$.
For problem $1.2.14$
<h3>Solution</h3>
<p>
Acceleration depends on time, as:
Missing or unrecognized delimiter for \left \left{\begin{matrix} a(t) = 0\;m/s^2,\; 0\text{ s} \leq t \leq 2\text{ s}\ a(t) = 20t,\; 2\text{ s} \leq t \leq 5\text{ s}\ a(t) = 60\;m/s^2,\; 5\text{ s} \leq t \leq 9\text{ s}\ a(t) = 60-20(t-9),\; 9\text{ s} \leq t \leq 12\text{ s}\ a(t) = 0, t> 12\text{ s} \end{matrix}\right. $$\left{\begin{matrix} a(t) = 0\;m/s^2,\; 0\text{ s} \leq t \leq 2\text{ s}\ a(t) = 20t,\; 2\text{ s} \leq t \leq 5\text{ s}\ a(t) = 60\;m/s^2,\; 5\text{ s} \leq t \leq 9\text{ s}\ a(t) = 60-20(t-9),\; 9\text{ s} \leq t \leq 12\text{ s}\ a(t) = 0, t> 12\text{ s} \end{matrix}\right.$$
Given that the area under the graph of acceleration vs. time is velocity vs. time. Then the velocity depends on time as:
Missing or unrecognized delimiter for \left \left{\begin{matrix} v(t) = 3\;m/s,\; 0\text{ s} \leq t \leq 2\text{ s}\ v(t) = 3+20t^2/2,\; 2\text{ s} \leq t \leq 5\text{ s}\ v(t) = 93+60(t-5),\; 5\text{ s} \leq t \leq 9\text{ s}\ v(t) = 333 - 20 (t - 9)^2/2 + 60 (t - 9),\; 9\text{ s} \leq t \leq 12\text{ s}\ v(t) = 423\;m/s, t> 12\text{ s} \end{matrix}\right. $$\left{\begin{matrix} v(t) = 3\;m/s,\; 0\text{ s} \leq t \leq 2\text{ s}\ v(t) = 3+20t^2/2,\; 2\text{ s} \leq t \leq 5\text{ s}\ v(t) = 93+60(t-5),\; 5\text{ s} \leq t \leq 9\text{ s}\ v(t) = 333 - 20 (t - 9)^2/2 + 60 (t - 9),\; 9\text{ s} \leq t \leq 12\text{ s}\ v(t) = 423\;m/s, t> 12\text{ s} \end{matrix}\right.$$
The graph of this dependence is presented below:
Dependence of velocity on time
It follows that at $4$ $15\text{ s}$ $43$ $423\\;m/s$
<h4>Answer</h4>
<p>
$$v_1=43\\;m/s; \\;v_2=423\\;m/s.$$
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